A body of 20 g and \( +50\,\mu\mathrm{C} \) charge is suspended from the ceiling by a string. If a horizontal electric field of 10 kV/m is applied, the angle between the string and vertical direction is (Take \( g = 10\,\mathrm{N/kg} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \tan^{-1}(2.50) \)
Three forces balance on the body: its weight \( mg \) downward, the electric force \( qE \) horizontally and the tension along the string. Resolving the tension into vertical and horizontal parts gives
\( \displaystyle T\cos\theta = mg \qquad T\sin\theta = qE \;\Rightarrow\; \tan\theta = \frac{qE}{mg} \)
\( m = 20\,\mathrm{g} = 0.02\,\mathrm{kg} \)
\( q = 50\,\mu\mathrm{C} = 5\times10^{-5}\,\mathrm{C} \)
\( E = 10\,\mathrm{kV/m} = 10^{4}\,\mathrm{N/C} \)
\( \displaystyle qE = 5\times10^{-5} \times 10^{4} = 0.5\,\mathrm{N} \qquad mg = 0.02 \times 10 = 0.2\,\mathrm{N} \)
\( \displaystyle \tan\theta = \frac{0.5}{0.2} = 2.5 \;\Rightarrow\; \theta = \tan^{-1}(2.50) \)